Question

the figure shown is composed of two cubes. which equation shows the surface area of the figure? s.a. = 6(3·3) + 6(2·2) - 2(2·2) s.a. = 6(3·3) + 6(2·2) + 2(2·2) s.a. = 6(3·3) + 6(2·2) - 2(3·3) s.a. = 6(3·3) + 6(2·2) + 2(3·3)

Explanation:

Step1: Recall surface area of a cube

The surface area of a single cube with side length \( s \) is \( 6s^2 \), which can be written as \( 6(s \cdot s) \). For the larger cube (side 3 mm), its surface area is \( 6(3 \cdot 3) \), and for the smaller cube (side 2 mm), it is \( 6(2 \cdot 2) \).

Step2: Account for overlapping area

When the two cubes are joined, the overlapping area (where they are attached) is covered twice initially (once in each cube's surface area) but should only be counted once. So we need to subtract twice the area of the overlapping face (since we added it twice, we subtract 2 times that area to get the correct total). The overlapping face has an area of \( 2 \cdot 2 \) (since the smaller cube's face is \( 2 \times 2 \)). So we subtract \( 2(2 \cdot 2) \) from the sum of the two cubes' surface areas.

Putting it together, the total surface area \( S.A. = 6(3 \cdot 3) + 6(2 \cdot 2) - 2(2 \cdot 2) \).

Answer:

\( \text{S.A.} = 6(3 \cdot 3) + 6(2 \cdot 2) - 2(2 \cdot 2) \) (the first option)